optical fibres

Discussion in 'Homework Help' started by bondpiki, Jul 23, 2014.

  1. bondpiki

    Thread Starter New Member

    Jul 23, 2014
    If receiver is a transimpedance amplifier with an open loop gain of 200 and the average amplifier noise current is 2.7 pA/
    Hz^0.5. The input resistance of the amplifier is 27 kΩ and the feedback resistance is 56 kΩ. The junction capacitance of the photodiode is 10 pF. Determine the photocurrent for the diode to maintain the required SNR of 50dB
    Last edited: Jul 23, 2014
  2. crutschow


    Mar 14, 2008
    Sorry, we don't do your homework for you. :rolleyes: But if you show us what you've done and where you have questions we may be able to help.
  3. Renesh Thakoordeen

    New Member

    Sep 15, 2015
    Hey I've got the same question as OP. What I've done is I've used two equations:

    B = G/(2*pi*Rf*C)

    Ip = (2*e*I*B)^0.5

    B - Bandwidth
    Rf- Feedback resistance
    C - capacitance. Although the equation says to use total capacitance. I'm not sure about Junction capacitance. Help?
    I - Average amplifier noise current
    G - Gain

    Let me know what you think.
  4. tvstar69

    New Member

    Sep 18, 2017
    Hi did this formula work out correct for the question?